Charges are placed on the vertices of a square as shown. Let $E$ be the electric field and $V$ the potential at the centre. If the charges on $A$ and $B$ are interchanged with those on $D$ and $C$ respectively, then 

A charge is spread non-uniformly on the surface of a hollow sphere of radius $R$, such that the charge density is given by $\sigma=\sigma_0(1-\sin \theta)$, where $\theta$ is the usual polar angle. The potential at the centre of the sphere is

Consider two charged metallic spheres $S_{1}$ and $\mathrm{S}_{2}$ of radii $\mathrm{R}_{1}$ and $\mathrm{R}_{2},$ respectively. The electric $\left.\text { fields }\left.\mathrm{E}_{1} \text { (on } \mathrm{S}_{1}\right) \text { and } \mathrm{E}_{2} \text { (on } \mathrm{S}_{2}\right)$ on their surfaces are such that $\mathrm{E}_{1} / \mathrm{E}_{2}=\mathrm{R}_{1} / \mathrm{R}_{2} .$ Then the ratio $\left.\mathrm{V}_{1}\left(\mathrm{on}\; \mathrm{S}_{1}\right) / \mathrm{V}_{2} \text { (on } \mathrm{S}_{2}\right)$ of the electrostatic potentials on each sphere is 

Consider two points $1$ and $2$ in a region outside a charged sphere. Two points are not very far away from the sphere. If $E$ and $V$ represent the electric field vector and the electric potential, which of the following is not possible

Assertion: Electron move away from a region of higher potential to a region of lower potential.

Reason: An electron has a negative charge.